THE STATE UNIVERSITY OF NEW JERSEY RUTGERS. College of Engineering Department of Electrical and Computer Engineering

Size: px

Start display at page:

Download "THE STATE UNIVERSITY OF NEW JERSEY RUTGERS. College of Engineering Department of Electrical and Computer Engineering"

Shannon West
5 years ago
Views:

1 THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Spring Problem Set 3 1. Discovered Angle Modulation A signal s(t) is measured and found to be described by s(t) = A cos(2πf a t + α sin 2πf b t) (a) We re later told that s(t) is an angle modulated signal with sensitivity k p. Using the standard angle modulation signal format found in your text, what is the information signal m(t)? SOLUTION: A phase-modulated signal s(t) is a form of angle modulation in where the angle θ i (t) is varied linearly with the message, thus this is described in time domain by s(t) = A cos(2πf c t + k p m(t)) where Using the above equations, we see that θ(t) = 2πf c t + k p m(t) θ(t) = 2πf a t + α sin 2πf b t Hence, f c = f a k p = α m(t) = sin(2πf b t) with f m = f b (b) Now, imagine that you re told WHOOPS! I meant FREQUENCY modulation. with frequency sensitivity k f. Again using the standard signal format described in your text, please provide the information signal m(t) and the instantaneous frequency f i (t). SOLUTION: A frequency-modulated signal s(t) is a form of angle modulation in which the instantenous frequency f i (t) is varied linearly with the message signal m(t), and it is given by t ) s(t) = A cos (2πf c t + 2πk f m(τ)dτ 0 where the instantenous frequency is defined as f i (t) = fc + k f m(t) = 1 dθ i (t) 2π dt 1

2 Hence, given our signal s(t) we find that θ i (t) = 2πf a t + α sin(2πf b t) and Therefore, dθ i (t) dt = 2πf a + α cos(2πf b t)(2πf b ) f i (t) = 1 dθ i (t) = f a + αf b cos(2πf b t) 2π dt f c = f a k f = αf b m(t) = cos(2πf b t) 2. Two Tone Madness: Consider a message signal with two tones at frequencies f a and f b respectively, defined as m(t) = A m cos(2πf a t) cos(2πf b t) (a) Find the corresponding phase modulated and frequency modulated signals. SOLUTION: For a PM signal, we have s(t) = A cos(2πf c t + k p m(t)) = A cos(2πf ( c t + k p A m cos(2πf a t) cos(2πf b t)) ) = A cos 2πf c t + kpam [cos(2π(f 2 a f b )t) + cos(2π(f a + f b )t)] For a FM signal, we have ( ) t s(t) = A cos 2πf c t + 2πk f m(τ)dτ 0 ( ) t = A cos 2πf c t + πk f A m [cos(2π(f 0 a f b )τ) + cos(2π(f a + f b )τ)]dτ ( [ ]) = A cos 2πf c t + k fa m sin(2π(fa f b )t) 2 (f a f b + sin(2π(fa+f b)t) ) (f a+f b ) (b) Find the narrowband FM (i.e NBFM) signal using the FM modulated signal obtained in the previous part. SOLUTION: Given the FM signal obtained in Part (a), we can obtain a narrowband FM signal by using a modulation index β for which it is very small compared to one radian. Thus, the FM signal is where s(t) = A cos ( 2πf c t + β [ sin(2π(fa f b )t) (f a f b ) β = A mk f 2 We use trigonometric expansions to obtain ( [ sin(2π(fa f b )t) s(t) = A cos (2πf c t)) cos β (f a f b ) 2 + sin(2π(f ]) a + f b )t) (f a + f b ) + sin(2π(f ]) ( [ a + f b )t) sin A sin (2πf c t)) sin β (f a + f b )

3 so the NBFM signal is approximately [ sin(2π(fa f b )t) s(t) = A cos(2πf c t) + Aβ sin(2πf c t) (f a f b ) since for small x 1 we have cos(x) 1 and sin(x) x + sin(2π(f ] a + f b )t) (f a + f b ) 3. Linear? Nonlinear? Let m 1 (t) and m 2 (t) be two message signals, and let s 1 (t) and s 2 (t) be the corresponding modulated signals. (a) Carefully show that if the modulation is DSB-SC, SSB, or VSB, then will produce a modulated signal m(t) = m 1 (t) + m 2 (t) s(t) = s 1 (t) + s 2 (t) SOLUTION: For m 1 (t) the modulated signal using DSB-SC is s 1 (t) = m 1 (t) cos(2πf c t) Similarly, for m 2 (t) using same modulation technique, we get and for m(t), we have Hence, s 2 (t) = m 2 (t) cos(2πf c t) s(t) = m(t) cos(2πf c t) = [m 1 (t) + m 2 (t)] cos(2πf c t) s(t) = s 1 (t) + s 2 (t) = [m 1 (t) + m 2 (t)] cos(2πf c t) Therefore, DSB-SC modulation is a linear modulation. Since SSB and VSB are simply linear operations (filtering) on this linear system, they are linear as well. (b) Show that if the modulation is PM or FM, then will not in general produce SOLUTION: For PM, we have and Thus, m(t) = m 1 (t) + m 2 (t) s(t) = s 1 (t) + s 2 (t) s 1 (t) = A cos(2πf c t + k p m 1 (t)) s 2 (t) = A cos(2πf c t + k p m 2 (t)) s(t) = A cos(2πf c t + k p m(t)) = A cos(2πf c t + k p [m 1 (t) + m 2 (t)]) s 1 (t) + s 2 (t) So, we notice that the phase is linear in the program material m i (t) but the overall signal is not. For FM, integration is a linear operation we use the results from PM to see that the instantaneous frequency and phase are linear but the actual modulated signal is not. 3

4 4. Lecture Redux Suppose we have a modulated signal with β << 1 (i.e., narrowband FM/PM). s(t) = A cos(2πf c t + βsin(2πf m t)) (a) Find the spectrum of this narrowband FM/PM signal. SOLUTION: Using the hint given in question 2, we can approximate the NBFM signal to be s(t) = A cos(2πf c t) + Aβ sin(2πf m t) sin(2πf c t) = A cos(2πf c t) + Aβ 2 (cos(2π(f c f m )t) cos(2π(f c + f m )t)) Hence, the spectrum of s(t) is S(f) = A 2 [δ(f f c) + δ(f + f c )] + Aβ 4 [δ(f (f c f m )) δ(f (f c + f m ))] (b) Compare your previous result to the spectrum of an AM (suppressed carrier) signal Cite similarities and differences. SOLUTION: and s(t) = A sin(2πf m t) cos(2πf c t) s(t) = A 2 (sin(2π( f c + f m )t) + sin(2π(f c + f m )t)) S(f) = A 4j [δ(f ( f c + f m )) + δ(f (f c + f m )) 2δ(f + (f c + f m ))] The AM signal is missing the large carrier present in the NBFM signal and the information is carried in phase with the carrier. IN the NBFM signal the information is π/2 out of phase with the carrier. Thus, the envelope of the NBFM signal is more or less constant while (as its name implies) the AM signal varies in amplitude in direct proportion to the program material. 5. Phase Locked Loops: Consider a phase locked loop whose sole purpose is to lock on to the incoming sinusoid cos(2πf c t) and produce sin(2πf c t) at the output of the VCO (voltage controlled oscillator). The incoming sinusoid is multiplied by the output of the VCO and the result is sent through a low pass filter whose output is multiplied by K (K > 0 a constant). The output of the multiplier, call it φ(t), is in turn sent to the input of the VCO which produces sin φ(t). (a) Sketch a system diagram of the PLL and show it will produce the desired result if the gain K is large enough. You may assume that φ(t) 2πf c t and then make appropriate approximations. SOLUTION: The inputs to the multiplier are cos(2πf c t) and sin φ(t). We rewrite φ(t) = 2πf c t+e(t) where e(t) is assumed small. At the output of the multiplier we then 4

5 have (after expanding the argument of the sine cos 2 (2πf c t) sin e(t)+cos(2πf c t) sin(2πf c t) and the second term is lost when it passes through the low pass filter since it s HIGH frequency (sin(4πf c t)/2 to be exact). The first term has a D.C. component of 1/2 (because of the cosine squared) and what pops out of the filter is e(t)/2 since sin x x for small x. We multiply this by K/2 and then have φ(t) = Ke(t)/2 = K(φ(t) 2πf c t)/2 which we rewrite as φ(t) + (K/2)φ(t) = Kπf c t This is a nice happy and stable first order differential equation whose homogeneous portion settles as e K 2 t which for K large is FAST FAST FAST! So given the homogeneous solution dies out, we re left with the particular for K large of φ(t) 2πf c t which is exactly what we wanted in the first place! If you made it to here, you solved the problem and were done. The following is extra. Now, above we assumed e(t) was small, but what if it s not small?!?! Specifically, what if e(t) = nπ + (t) where (t) IS small but n can be ANY INTEGER!!!. All the same approximations hold, except that what pops out of the LPF is ± (t), not e(t). We then have φ(t) = ±K (t)/2 = ±K(φ(t) 2πf c t n2π)/2 and we have lock again if n is even (just with a nπ phase offset). However, if n is odd, then what pops out of the sine is MINUS and that leads to INSTABILITY because the homogeneous equation will have a POSITIVE exponent. But that can t last long because the instability will make φ(t) grow so that we move up to the next value of n. So, any n odd gets kicked to the nearest n even solution. So the PLL is all about STABILITY! This system LOCKS ON to the input sinusoid BECAUSE it s a stable system (stable differential equation). The negative feedback stabilizes things just like negative feedback stabilizes op-amp circuits. Negative feedback is NEAT! (b) Given the same input cos(2πf c t) is it possible for the phase locked loop to lock on to frequencies other than f c. If so, which ones? If not, why not? HINT: Think about the bandwidth of the low pass filter which we never actually specified in the previous part. SOLUTION: Let s start with the previous part, but not assume e(t) is NOT small. The input to the low pass filter is cos 2 (2πf c t) sin e(t) + cos(2πf c t) sin(2πf c t) Certainly we re gonna lose the really high frequency second term and the cosine squared term averages to 1/2 so φ(t) = 1 sin e(t) 2 Now, the question becomes how low is low on the low pass filter? 5

6 Let s do some thought experiments. Suppose the PLL (that s technicalese for phase locked loop) frequency starts out too low (f 0 f c ). Then e(t) is large and φ = 0 because sin e(t) never makes it out of the LPF. φ = 0 implies zero frequency so the VCO output goes even lower until it s a D.C. level. No lock! We can make the same argument if the VCO starts too high in frequency too. So, there s a RANGE of frequencies which can be locked in. Let s say the low pass filter has bandwidth W Hz (single sided). Then we know that de(t) < 2πW to achieve lock. dt This means that the initial VCO frequency f 0 has to f 0 f c < W Otherwise the VCO outputs whatever it s lowest frequency is and thinks it s doing a good job! If you re REALLY interested in PLLs, our local expert is Dave Daut. He can tell you as much as (or more than) you ll ever want to know about PLLs since he was heavy into the topic when it was hot quite some time ago and still teaches the rudiments in his senior level course on RF engineering. The down side is that PLLs were studied to death quite a long time ago and are essentially a dead research topic. But so are vacuum tube audio amplifiers and they re still fun to learn about and use. 6

Solution to Chapter 4 Problems

Solution to Chapter 4 Problems Problem 4.1 1) Since F[sinc(400t)]= 1 modulation index 400 ( f 400 β f = k f max[ m(t) ] W Hence, the modulated signal is ), the bandwidth of the message signal is W = 00